Question Number 42521 by rahul 19 last updated on 27/Aug/18
$$\mathrm{Let}\:\overset{\rightarrow\:} {\mathrm{a}},\:\overset{\rightarrow} {\mathrm{b}}\:,\:\overset{\rightarrow} {\mathrm{c}}\:\mathrm{be}\:\mathrm{three}\:\mathrm{unit}\:\mathrm{vectors} \\ $$$$\mathrm{such}\:\mathrm{that}\:\mathrm{3}\overset{\rightarrow} {\mathrm{a}}+\mathrm{4}\overset{\rightarrow} {\mathrm{b}}+\mathrm{5}\overset{\rightarrow} {\mathrm{c}}\:=\:\mathrm{0}.\:\mathrm{Then}\:\mathrm{prove} \\ $$$$\mathrm{that}\:\overset{\rightarrow\:} {\mathrm{a}},\:\overset{\rightarrow} {\mathrm{b}},\overset{\rightarrow} {\mathrm{c}}\:\mathrm{are}\:\mathrm{coplanar}. \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 27/Aug/18
$${we}\:{have}\:{to}\:{prove}\:\left(\overset{\rightarrow} {{a}}×\overset{\rightarrow} {{b}}\right).\overset{\rightarrow} {{c}}=\mathrm{0} \\ $$$$\mathrm{3}\overset{\rightarrow} {{a}}+\mathrm{5}\overset{\rightarrow} {{c}}=−\mathrm{4}\overset{\rightarrow} {{b}} \\ $$$$\left(\mathrm{3}\overset{\rightarrow} {{a}}+\mathrm{5}\overset{\rightarrow} {{c}}\right)×\overset{\rightarrow} {{b}}=−\mathrm{4}\overset{\rightarrow} {{b}}×\overset{\rightarrow} {{b}} \\ $$$$\mathrm{3}\overset{\rightarrow} {{a}}×\overset{\rightarrow} {{b}}+\mathrm{5}\overset{\rightarrow} {{c}}×\overset{\rightarrow} {{b}}=\mathrm{0} \\ $$$$\mathrm{3}\overset{\rightarrow} {{a}}×\overset{\rightarrow} {{b}}=−\mathrm{5}\overset{\rightarrow} {{c}}×\overset{\rightarrow} {{b}} \\ $$$$\mathrm{3}\left(\overset{\rightarrow} {{a}}×\overset{\rightarrow} {{b}}\right).\overset{\rightarrow} {{c}}=−\mathrm{5}\left(\overset{\rightarrow} {{c}}×\overset{\rightarrow} {{b}}\right).\overset{\rightarrow} {{c}} \\ $$$${hence} \\ $$$$\mathrm{3}\left(\overset{\rightarrow} {{a}}×\overset{\rightarrow} {{b}}\right).\overset{\rightarrow} {{c}}=\mathrm{0}\:\:\:\:\:\:\:\:{so}\:\left(\overset{\rightarrow} {{a}}×\overset{\rightarrow} {{b}}\right).\overset{\rightarrow} {{c}}=\mathrm{0} \\ $$$${hdnce}\:{proved} \\ $$$$ \\ $$$$ \\ $$
Commented by rahul 19 last updated on 31/Aug/18
thanks sir.